Relation between a finite group and the Galois group for the field extension generated by the character table entries

Question

Let $E$ be the extension of $\mathbb{Q}$ generated by the character table entries of a finite group $G$.
Let $F$ be the Galois closure of $E$.

Examples:
- $G=C_n$, $E=F=\mathbb{Q}(\zeta_n)$ and $Gal(F/\mathbb{Q}) \simeq (\mathbb{Z}/n\mathbb{Z})^{\times} \simeq Out(G)=Aut(G)$,
- $G=S_n$, $E=F=\mathbb{Q}$ and so $C_1 = Gal(F/\mathbb{Q}) \hookrightarrow Out(G)$ and $Aut(G)$,
- $G=A_n$ with $3 \le n \le 5$, $E=F$ and $Gal(F/\mathbb{Q}) \simeq Out(G) \simeq C_2 \hookrightarrow Aut(G)$,
- $G=A_6$, $E=F=\mathbb{Q}(\sqrt{5})$ and $Gal(F/\mathbb{Q}) \simeq C_2 \hookrightarrow Out(G)\simeq C_2 \times C_2 \hookrightarrow Aut(G)$.

Question: Is it true in general that $Gal(F/\mathbb{Q})$ or $Aut(E/\mathbb{Q}) \hookrightarrow Out(G)$ or $Aut(G)$?
If not, is there at least a non-trivial homomorphism from $Gal(F/\mathbb{Q})$ to $Aut(G)$? Or any relation?

Bonus question: Is there an example with $E \subsetneq F$?

Answer 1

For the Mathieu group $G=M_{11}$, we have $E=\mathbb{Q}(i\sqrt{2},i\sqrt{11})$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/\mathbb{Q}) \not \hookrightarrow Out(G)$.

Otherwise the following sentence comes from this page:

The outer automorphism group acts on the character table by permuting columns (conjugacy classes) and accordingly rows, which gives another symmetry to the table.